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The work done when two mole of an ideal gas is compressed form a volume of `5 m^(3)` to `1 dm^(3)` at 300 K , under a pressure of 100 kPa isA. `499.9` kJB. `-499.9 `kJC. `-99.5` kJD. `42495`kJ |
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Answer» Correct Answer - A Work done , W = `- P_(ext") (V_(2) - V_(1))` Here , `P_("ext") = 100 k Pa` `V_(1) = 5 m^(3) = 5 xx 10^(3) L implies V_(2) = 1 dm^(3) = 1 L` `therefore W = -100 xx (1 - 500)` ` = - 499900 k PaL` `= - 499900 J = - 499.9 kJ ( because 1 k Pa L = 1 J)` |
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